Problem: Find $\lim_{\theta\to\scriptsize\dfrac{\pi}{2}}\dfrac{\sin^2(2\theta)}{1-\sin^2(\theta)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $2$ (Choice C) C $4$ (Choice D) D The limit doesn't exist
Solution: Substituting $\theta=\dfrac{\pi}{2}$ into $\dfrac{\sin^2(2\theta)}{1-\sin^2(\theta)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\sin(2\theta)$ in our expression, let's rewrite it using its double angle identity $\sin(2\theta)=2\sin(\theta)\cos(\theta)$. $\begin{aligned} &\phantom{=}\dfrac{\sin^2(2\theta)}{1-\sin^2(\theta)} \\\\ &=\dfrac{4\sin^2(\theta)\cos^2(\theta)}{1-\sin^2(\theta)} \gray{\sin(2\theta)\text{ identity}} \\\\ &=\dfrac{4\sin^2(\theta)(1-\sin^2(\theta))}{1-\sin^2(\theta)} \gray{\text{The Pythagorean identity}} \\\\ &=\dfrac{4\sin^2(\theta)(\cancel{1-\sin^2(\theta)})}{\cancel{(1-\sin^2(\theta))}} \gray{\text{Cancel common factors}} \\\\ &=4\sin^2(\theta)\text{, for }\theta\neq \{...,- \dfrac{7\pi}{2}, - \dfrac{3\pi}{2}, \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2},...\} \end{aligned}$ This means that the two expressions have the same value for all $\theta $ -values (in their domains) except for $(2k+1)\dfrac{\pi}{2}$ for any integer $k$, and specifically $\dfrac{\pi}{2}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\sin^2(2\theta)}{1-\sin^2(\theta)}=4\sin^2(\theta)$ for all $\theta$ -values in the interval $(-\pi,\pi)$ except for $\theta=\dfrac{\pi}{2}$. Therefore, $\lim_{\theta\to \scriptsize\dfrac{\pi}{2}}\dfrac{\sin^2(2\theta)}{1-\sin^2(\theta)}=\lim_{\theta\to \scriptsize\dfrac{\pi}{2}}4\sin^2(\theta)=4$ (The last limit was found using direct substitution.) In conclusion, $\lim_{\theta\to\scriptsize\dfrac{\pi}{2}}\dfrac{\sin^2(2\theta)}{1-\sin^2(\theta)}=4$.